Question

A boy has 20% chance of hitting at a target. Let, $p$ denote the probability of hitting the target for the first time at the $n^{th}$ trial. If $p$ satisfies the inequality $625p^2-175p+12<0$, then value of n is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is C $3$

$p\left(n\right)$ = Hitting for the 1st time at the n-th trial

$⟹p\left(n\right)=(0.8{)}^{n-1}.0.2$

Boundary of the inequality is the solution of a quadratic equation in p: $625p^2-175p+12=0$

disc: ${175}^2-4\times 12\times 625=625={25}^2$

$⟹p=\frac{175\pm 25}{1250}=\frac{3}{25}$ or $\frac{4}{25}$

So, $p\left(n\right)$ is negative between those two values.

$p\left(n\right)=\frac{3}{25}=(0.8{)}^{n-1}.0.2$

$⟹\frac{3}{5}=(0.8{)}^{n-1}$

$⟹\log \left(dfrac35\right)=(n-1)\log \left(0.8\right)$

$⟹n=1+\frac{\log \left(dfrac35\right)}{\log \left(0.8\right)}=3.289$

$p\left(n\right)=\frac{4}{25}=(0.8{)}^{n-1}.0.2$

$⟹\frac{4}{5}=(0.8{)}^{n-1}$

$⟹n=1+\frac{\log \left(dfrac45\right)}{\log \left(0.8\right)}=2$

So, $2<n<\mathrm{3.289..}⟹n=3$ (as n is integer)